Question

Let ABC be a triangle with circumcircle Γ and incentre I. Let the internal angle bisectors of ∠A, ∠B and ∠C meet Γ in A', B' and C' respectively. Let B'C' intersect AA' in P and AC in Q, and let BB0 intersect AC in R. Suppose the quadrilateral PIRQ is a kite; that is, IP = IR and QP = QR. Prove that ABC is an equilateral triangle.

Answer 1

We rst show that AA' is perpendicular to B'C'. Observe ∠C'A'A = ∠C'CA = ∠C/2, ∠A'C'C = ∠A'AC = ∠A/2, and ∠CC'B'= ∠CBB' = ∠B/2. Hence

It follows that ∠APC'= ∠A'PC' = 90⁰. Thus ∠IPQ = 90⁰. Since PIRQ is a kite, we observe that ∠IPR = ∠IRP and ∠QPR = ∠QRP. This implies that ∠IRQ = 90⁰. Hence the kite IRQP is also a cyclic quadrilateral. Since ∠IRQ = 90⁰, we see that BB' ⊥ AC. Since BB' is the bisector of ∠B, we conclude that ∠A = ∠C.

We also observe that the triangles IRC and IPB' are congruent triangles: they are similar, since ∠IRC = ∠IPB' = 90⁰ and ∠ICR = ∠C=2 = ∠IB'P(= ∠BCC'), besides IR = IP. Therefore IC = IB'. But B'I = B'C. Thus IB'C is an equilateral triangle. This means ∠B'IC = 60⁰ and hence ∠ICR = 30⁰. Therefore ∠C=2 = 30⁰. Hence ∠A = ∠C = 60⁰. It follows that ABC is equilateral.